Question 1

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s².At the end of 2.0 s its angular momentum about the origin is: A)0 kg∙m²/s B)(-36 kg∙m²/s) C)(+48 kg∙m²/s) D)(-96 kg∙m²/s) E)(+96 kg∙m²/s)

Accepted Answer

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Question 2

The fundamental dimensions of angular momentum are: A)mass·length·time^-1 B)mass·length^-2·time^-2 C)mass²·time^-1 D)mass·length²·time^-2 E)none of these

Accepted Answer

The answer of The fundamental dimensions of angular momentum are: A)mass·length·time^-1 B)mass·length^-2·time^-2 C)mass²·time^-1 D)mass·length²·time^-2 E)none...

Question 3

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The component in the xy plane of the angular momentum around the origin has magnitude: A)0 kg∙m²/s B)6.0 kg∙m²/s C)9.0 kg∙m²/s D)11 kg∙m²/s E)14 kg∙m²/s

Accepted Answer

The answer of A 2.0-kg block travels around a 0.50-m...

Question 4

A 15-g paper clip is attached to the rim of a phonograph record with a diameter of 30 cm, spinning at 3.5 rad/s.The magnitude of its angular momentum is: A)1.2 *10^-3 kg∙m²/s B)4.7 *10^-3 kg∙m²/s C)7.9 * 10^-³ kg∙m²/s D)1.6*10^-² kg∙m²/s E)1.2 kg∙m²/s

Accepted Answer

The answer of A 15-g paper clip is attached to...

Question 5

A 6.0-kg particle moves to the right at 4.0 m/s as shown.The magnitude of its angular momentum about the point O is: A)0 kg∙m²/s B)288 kg∙m²/s C)144 kg∙m²/s D)24 kg∙m²/s E)249 kg∙m²/s

Accepted Answer

The answer of A 6.0-kg particle moves to the right...

Question 6

Two objects are moving in the x, y plane as shown.The magnitude of their total angular momentum (about the origin O)is: A)0 kg∙m²/s B)6 kg∙m²/s C)12 kg∙m²/s D)30 kg∙m²/s E)78 kg∙m²/s

Accepted Answer

The answer of Two objects are moving in the x,...

Question 7

The angular momentum vector of Earth, due to its daily rotation, is directed:
A)tangent to the equator toward the east
B)tangent to the equator toward the west
C)north
D)south
E)toward the sun

Accepted Answer

The angular momentum vector of Earth due to its daily rotation is directed north because it points perpendicular to the plane of rotation, which, for Earth's rotation around its axis, is along the axis from the south pole to the north pole.

Question 8

Possible units of angular momentum are: A)kg.m/s B)kg.m²/s² C)kg.m/s² D)kg.m²/s E)none of these

Accepted Answer

The units of angular momentum are kg·m²/s. The given option D is the only one that has these units.

Question 9

A particle moves along the x axis.In order to calculate the torque on the particle, you need to know:
A)the velocity of the particle
B)the rotational inertia of the particle
C)the point about which the torque is to be calculated
D)the kinetic energy of the particle
E)the mass of the particle

Accepted Answer

Torque is defined as the cross product of the lever arm (distance from the pivot point to the point of force application) and the force applied. Therefore, knowing the point about which the torque is to be calculated is essential to determine the lever arm distance, which is crucial for calculating torque. The other options, such as velocity, rotational inertia, kinetic energy, and mass, do not directly influence the calculation of torque without considering the force applied and the lever arm distance from a specific point.

Question 10

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75 m from the origin.The z component of the angular momentum around the origin is: A)6.0 kg∙m²/s B)9.0 kg∙m²/s C)11 kg∙m²/s D)14 kg∙m²/s E)20 kg∙m²/s

Accepted Answer

The answer of A 2.0-kg block travels around a 0.50-m...

Question 11

The newton.second is a unit of:
A)work
B)angular momentum
C)power
D)linear momentum
E)none of these

Accepted Answer

The answer of The newton.second is a unit of:
A)work
B)angular momentum
C)power
D)linear...

Question 12

The unit kg.m²/s can be used for: A)angular momentum B)rotational kinetic energy C)rotational inertia D)torque E)power

Accepted Answer

The answer of The unit kg.m²/s can be used for: A)angular...

Question 13

A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s.The magnitude of its angular momentum about the center of the circle is: A)6.0 kg∙m²/s B)12 kg∙m²/s C)48 kg∙m²/s D)72 kg∙m²/s E)576 kg∙m²/s

Accepted Answer

The answer of A 2.0-kg block travels around a 0.50-m...

Question 14

A single force acts on a particle situated on the positive x axis.The torque about the origin is in the negative z direction.The force might be:
A)in the positive y direction
B)in the negative y direction
C)in the positive x direction
D)in the negative x direction
E)in the positive z direction

Accepted Answer

The torque about the origin is given by the cross product of the position vector and the force: 
$\vec{	au} = \vec{r} 	imes \vec{F}$ 
Since the torque is in the negative z direction, this means that the force and position vector must be perpendicular to the z-axis. The only force that satisfies this condition is a force in the negative y direction, since the cross product of any vector in the positive x direction with a vector in the negative y direction will be in the negative z direction. Therefore, the force must be in the negative y direction.

Question 15

A particle moves along the x axis.In order to calculate the angular momentum of the particle, you need to know:
A)the size of the particle
B)the rotational inertia of the particle
C)the point about which the angular momentum is to be calculated
D)the kinetic energy of the particle
E)the acceleration of the particle

Accepted Answer

Angular momentum is calculated with respect to a specific point or axis, so it is necessary to know the point about which the angular momentum is to be calculated. The other choices are not directly related to calculating angular momentum.

Question 16

A particle is located on the x axis at x = 2.0 m from the origin.A force of 25 N, directed 30° above the x axis in the x-y plane, acts on the particle.What is the torque about the origin on the particle?
A)50 N∙m, in the positive z direction
B)25 N∙m, in the positive z direction
C)50 N∙m, in the negative z direction
D)25 N∙m, in the negative z direction
E)There is no torque about the origin.

Accepted Answer

The answer of A particle is located on the x...

Question 17

Which of the following is NOT a vector?
A)linear momentum
B)angular momentum
C)rotational inertia
D)torque
E)angular velocity

Accepted Answer

Rotational inertia is a scalar quantity, not a vector. It is a measure of an object's resistance to changes in rotational motion. The other choices listed are all vectors.

Question 18

A force  = 4.2 N + 3.7 N + 1.2 N acts on a particle located at x = 3.3 m.What is the torque on the particle around the origin?
A)14 N∙m
B)-4.0 N∙m + 12 N∙m
C)12 N∙m
D)14 N∙m - 4.0 N∙m + 12 N∙m
E)cannot be calculated without knowing the mass of the particle

Accepted Answer

The answer of A force  = 4.2 N +...

Question 19

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s².The torque, relative to the origin, acting on it at the end of 2.0 s is: A)0 N∙m B)(-18 N∙m) C)(+24 N∙m) D)(-144 N∙m) E)(+144 N∙m)

Accepted Answer

The answer of A 2.0-kg block starts from rest on...

Question 20

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The magnitude of its angular momentum around the origin is: A)6.0 kg∙m²/s B)9.0 kg∙m²/s C)11 kg∙m²/s D)14 kg∙m²/s E)20 kg∙m²/s

Accepted Answer

The formula for angular momentum is L = Iω, where I is the moment of inertia and ω is the angular speed.

We can find the moment of inertia of the block using the formula for the moment of inertia of a uniform solid sphere rotating about an axis passing through its center:

I = (2/5)MR^2

where M is the mass of the block and R is the radius of the circle. Plugging in the given values, we get:

I = (2/5)(2 kg)(0.5 m)^2 = 0.5 kg·m^2

Then, we can calculate the angular momentum using the formula:

L = Iω = (0.5 kg·m^2)(12 rad/s) = 6 kg·m^2/s

However, we need to find the magnitude of the angular momentum around the origin. Since the circle is centered on the z-axis and parallel to the xy-plane, the angular momentum vector is perpendicular to the xy-plane and has a magnitude equal to its z-component:

Lz = Iωz = Iω = (0.5 kg·m^2)(12 rad/s) = 6 kg·m^2/s

Therefore, the magnitude of the angular momentum around the origin is 6 kg·m^2/s, which is closest to choice C).

Quiz 11: Rolling, Torque, and Angular Momentum

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s².At the end of 2.0 s its angular momentum about the origin is:

The fundamental dimensions of angular momentum are:

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The component in the xy plane of the angular momentum around the origin has magnitude:

A 15-g paper clip is attached to the rim of a phonograph record with a diameter of 30 cm, spinning at 3.5 rad/s.The magnitude of its angular momentum is:

A 6.0-kg particle moves to the right at 4.0 m/s as shown.The magnitude of its angular momentum about the point O is:

Two objects are moving in the x, y plane as shown.The magnitude of their total angular momentum (about the origin O) is:

The angular momentum vector of Earth, due to its daily rotation, is directed:

Possible units of angular momentum are:

A particle moves along the x axis.In order to calculate the torque on the particle, you need to know:

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75 m from the origin.The z component of the angular momentum around the origin is:

The newton.second is a unit of:

The unit kg.m²/s can be used for:

A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s.The magnitude of its angular momentum about the center of the circle is:

A single force acts on a particle situated on the positive x axis.The torque about the origin is in the negative z direction.The force might be:

A particle moves along the x axis.In order to calculate the angular momentum of the particle, you need to know:

A particle is located on the x axis at x = 2.0 m from the origin.A force of 25 N, directed 30° above the x axis in the x-y plane, acts on the particle.What is the torque about the origin on the particle?

Which of the following is NOT a vector?

A force = 4.2 N + 3.7 N + 1.2 N acts on a particle located at x = 3.3 m.What is the torque on the particle around the origin?

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s².The torque, relative to the origin, acting on it at the end of 2.0 s is:

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The magnitude of its angular momentum around the origin is:

Quiz 11: Rolling, Torque, and Angular Momentum

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s2.At the end of 2.0 s its angular momentum about the origin is:

The fundamental dimensions of angular momentum are:

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The component in the xy plane of the angular momentum around the origin has magnitude:

A 15-g paper clip is attached to the rim of a phonograph record with a diameter of 30 cm, spinning at 3.5 rad/s.The magnitude of its angular momentum is:

A 6.0-kg particle moves to the right at 4.0 m/s as shown.The magnitude of its angular momentum about the point O is:

Two objects are moving in the x, y plane as shown.The magnitude of their total angular momentum (about the origin O) is:

The angular momentum vector of Earth, due to its daily rotation, is directed:

Possible units of angular momentum are:

A particle moves along the x axis.In order to calculate the torque on the particle, you need to know:

A 2.0-kg block travels around a 0.50-m radius circle with an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, a distance of 0.75 m from the origin.The z component of the angular momentum around the origin is:

The newton.second is a unit of:

The unit kg.m2/s can be used for:

A 2.0-kg block travels around a 0.50-m radius circle with an angular velocity of 12 rad/s.The magnitude of its angular momentum about the center of the circle is:

A single force acts on a particle situated on the positive x axis.The torque about the origin is in the negative z direction.The force might be:

A particle moves along the x axis.In order to calculate the angular momentum of the particle, you need to know:

A particle is located on the x axis at x = 2.0 m from the origin.A force of 25 N, directed 30° above the x axis in the x-y plane, acts on the particle.What is the torque about the origin on the particle?

Which of the following is NOT a vector?

A force = 4.2 N + 3.7 N + 1.2 N acts on a particle located at x = 3.3 m.What is the torque on the particle around the origin?

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s2.The torque, relative to the origin, acting on it at the end of 2.0 s is:

As a 2.0-kg block travels around a 0.50-m radius circle it has an angular speed of 12 rad/s.The circle is parallel to the xy plane and is centered on the z axis, 0.75 m from the origin.The magnitude of its angular momentum around the origin is:

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s².At the end of 2.0 s its angular momentum about the origin is:

The unit kg.m²/s can be used for:

A 2.0-kg block starts from rest on the positive x axis 3.0 m from the origin and thereafter has an acceleration given by in m/s².The torque, relative to the origin, acting on it at the end of 2.0 s is: